arXivDaily arXiv每日学术速递 周一至周五更新
arXiv周末暂无论文更新,休息一下吧,周末愉快~~
arXiv 0901.2093math.NTmath.LO
收藏

具有有限个解的丢番图方程解的一个假设上界

A hypothetical upper bound for the solutions of a Diophantine equation with a finite number of solutions

Apoloniusz Tyszka

摘要:

我们猜想,如果系统S \subseteq {x_i=1,x_i+x_j=x_k,x_i \cdot x_j=x_k:i,j,k \in {1,.,n}}在整数x_1,.,x_n,则每个这样的解(x_1,...,x_n)满足|x_1|,.,|x_n| \leq 2^{2^{n-1}}。通过猜想,如果一个丢番图方程在整数(非负整数,有理数)中只有1/2个解,那么它们的高度由方程的次数和系数的可计算函数从上而下有界。这个猜想暗示了具有无穷多个整数(非负整数)解的丢番图方程组是递归可解的。对于任意可计算界而不是2^{2^{n-1}}的猜想仍然与Matiyasevich的猜想相矛盾,即每个递归可枚举集M \subseteq { \mathbb N}^n具有有限重丢番图表示。

英文摘要:

We conjecture that if a system S \subseteq {x_i=1, x_i+x_j=x_k, x_i \cdot x_j=x_k: i,j,k \in {1,...,n}} has only finitely many solutions in integers x_1,...,x_n, then each such solution (x_1,...,x_n) satisfies |x_1|,...,|x_n| \leq 2^{2^{n-1}}. By the conjecture, if a Diophantine equation has only finitely many solutions in integers (non-negative integers, rationals), then their heights are bounded from above by a computable function of the degree and the coefficients of the equation. The conjecture implies that the set of Diophantine equations which have infinitely many solutions in integers (non-negative integers) is recursively enumerable. The conjecture stated for an arbitrary computable bound instead of 2^{2^{n-1}} remains in contradiction to Matiyasevich's conjecture that each recursively enumerable set M \subseteq {\mathbb N}^n has a finite-fold Diophantine representation.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

↑